Compound A ($C_8H_{10}O$) is found to react with NaOI (produced by reacting Y with NaOH) and yield a yellow pr

Question

Compound A ($C_8H_{10}O$) is found to react with NaOI (produced by reacting Y with NaOH) and yield a yellow precipitate with a characteristic smell. A and Y are respectively?

Accepted Answer

1. **Identify the Reaction:** The formation of a yellow precipitate with a characteristic smell upon reaction with sodium hypoiodite (NaOI) is the specific confirmatory test for **Iodoform** ($CHI_3$). 
2. **Identify Reagent Y:** NaOI is prepared *in situ* by the reaction of **Iodine** ($I_2$) with Sodium Hydroxide (NaOH). Therefore, Y is Iodine. ($I_2 + 2NaOH \rightarrow NaOI + NaI + H_2O$) 
3. **Identify Compound A:** The iodoform test is positive for compounds containing the methyl ketone group ($CH_3-C=O$) or the methyl carbinol group ($CH_3-CH(OH)-$) linked to a carbon or hydrogen.
    *   The molecular formula is given as $C_8H_{10}O$.
    *   **1-Phenylethanol** ($C_6H_5-CH(OH)-CH_3$) fits this formula ($C_8H_{10}O$) and contains the required secondary alcohol group ($CH_3-CH(OH)-$) attached to a phenyl ring.
    *   Upon oxidation by NaOI, it converts to acetophenone and then undergoes the haloform reaction to yield iodoform ($CHI_3$) and sodium benzoate.
    *   *Note:* 2-Phenylethanol ($C_6H_5-CH_2-CH_2-OH$) does not have the required grouping. Acetophenone ($C_6H_5COCH_3$) reacts but its formula is $C_8H_8O$, not $C_8H_{10}O$.

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